Maharashtra Board Class 11 Maths Part 1 Chapter 5 Locus and Straight Line 5.3 Solutions

Question 1. Write the equation of the line:
(a) parallel to the X-axis and at a distance of 5 units from it and above it.
(b) parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
(c) parallel to the X-axis and at a distance of 4 units from the point (-2, 3).
Answer:
(a) Equation of a line parallel to the X-axis is y = k.
Since the line is at a distance of 5 units above the X-axis.
\( \therefore \) k = 5
\( \therefore \) the equation of the required line is y = 5.
(b) Equation of a line parallel to the Y-axis is x = h.
Since the line is at a distance of 5 units to the left of the Y-axis.
\( \therefore \) h = -5
\( \therefore \) the equation of the required line is x = -5.
(c) Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since, the line is at a distance of 4 units from the point (-2, 3).
\( \therefore \) k = 3 + 4 = 7 or k = 3-4 = -1
\( \therefore \) the equation of the required line is y = 7 or y = -1.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कार्तीय निर्देशांक प्रणाली को दर्शाता है जिसमें x-अक्ष और y-अक्ष हैं। इसमें दो क्षैतिज रेखाएँ, y=7 और y=-1, खींची गई हैं, जो x-अक्ष के समानांतर हैं। एक बिंदु (-2, 3) दर्शाया गया है, और रेखाएँ इस बिंदु के y-निर्देशांक (3) से 4 इकाई की दूरी पर स्थित हैं, ऊपर की ओर (y=7) और नीचे की ओर (y=-1) दोनों दिशाओं को दिखाते हुए।
In simple words: The equation of a line parallel to the X-axis is y=k, and parallel to the Y-axis is x=h. The constant k or h is determined by the given distance from the axis and its direction (above/below, left/right), or by the distance from a given point's coordinate.

🎯 Exam Tip: Remember the standard forms for lines parallel to axes. For a line parallel to the X-axis, its equation is y = k, and for a line parallel to the Y-axis, its equation is x = h. The sign of k or h depends on the side of the axis.

Question 2. Obtain the equation of the line:
(a) parallel to the X-axis and making an intercept of 3 units on the Y-axis.
(b) parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Answer:
(a) Equation of a line parallel to X-axis with y-intercept 'k' is y = k.
Here, y-intercept = 3
\( \therefore \) the equation of the required line is y = 3.
(b) Equation of a line parallel to Y-axis with x-intercept 'h' is x = h.
Here, x-intercept = 4
\( \therefore \) the equation of the required line is x = 4.
In simple words: The equation of a line parallel to an axis is directly its intercept value on the other axis. So, a line parallel to the X-axis with a y-intercept of 3 is y=3, and a line parallel to the Y-axis with an x-intercept of 4 is x=4.

🎯 Exam Tip: Intercepts directly define the constant in the equation of lines parallel to the coordinate axes. A y-intercept of 'k' means the line passes through (0,k), making its equation y=k if parallel to X-axis.

Question 3. Obtain the equation of the line containing the point:
(a) A(2, -3) and parallel to the Y-axis.
(b) B(4, -3) and parallel to the X-axis.
Answer:
(a) Equation of a line parallel to the Y-axis is of the form x = h.
Since, the line passes through A(2, -3).
\( \therefore \) h = 2
\( \therefore \) the equation of the required line is x = 2.
(b) Equation of a line parallel to the X-axis is of the form y = k.
Since, the line passes through B(4, -3)
\( \therefore \) k = -3
\( \therefore \) the equation of the required line is y = -3.
In simple words: If a line is parallel to the Y-axis and passes through a point (x, y), its equation is x = the x-coordinate of the point. Similarly, if it's parallel to the X-axis, its equation is y = the y-coordinate of the point.

🎯 Exam Tip: When a line is parallel to an axis and passes through a specific point, its equation is simply the constant coordinate corresponding to the axis it's parallel to. For example, x=constant for parallel to Y-axis, y=constant for parallel to X-axis.

Question 4. Find the equation of the line passing through the points A(2, 0) and B(3, 4).
Answer:
Solution:
The required line passes through the points A(2, 0) = \( (x_1, y_1) \) and B(3, 4) = \( (x_2, y_2) \) say.
Equation of the line in two-point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
\( \therefore \) the equation of the required line is
\( \frac{y-0}{4-0} = \frac{x-2}{3-2} \)
\( \therefore \frac{y}{4} = \frac{x-2}{1} \)
\( \therefore \) y = 4(x - 2)
\( \therefore \) y = 4x - 8
\( \therefore \) 4x - y - 8 = 0
Check:
If the points A(2, 0) and B(3, 4) satisfy 4x - y - 8 = 0, then our answer is correct.
For point A(2, 0),
L.H.S. = 4x - y - 8
= 4(2) – 0 – 8
= 8 - 8
= 0
= R.H.S.
For point B(3, 4),
L.H.S. = 4x - y - 8
= 4(3) – 4 – 8
= 12 - 12
= 0
= R.H.S.
Thus, our answer is correct.
In simple words: To find the equation of a line passing through two given points, use the two-point form formula \( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \). Substitute the coordinates of the two points and simplify the equation into the standard form.

🎯 Exam Tip: Always remember the two-point form of a linear equation. After finding the equation, a good practice is to substitute both given points back into the final equation to verify if they satisfy it, ensuring accuracy.

Question 5. Line y = mx + c passes through the points A(2, 1) and B(3, 2). Determine m and c.
Answer:
Solution:
Given, A(2, 1) and B(3, 2).
Equation of a line in two-point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
\( \therefore \) the equation of the passing through A and B line is
\( \frac{y-1}{2-1} = \frac{x-2}{3-2} \)
\( \therefore \frac{y-1}{1} = \frac{x-2}{1} \)
\( \therefore \) y - 1 = x - 2
\( \therefore \) y = x - 1
Comparing this equation with y = mx + c, we get
m = 1 and c = -1
Alternate method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
\( \therefore \) They must satisfy the equation.
\( \therefore \) 2m + c = 1 ........(i)
and 3m + c = 2 ......(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get
2(1) - c = 1
\( \therefore \) c = 1 - 2 = -1
In simple words: We can find the slope (m) and y-intercept (c) of a line by either using the two-point formula to get the line's equation and then comparing it to y=mx+c, or by substituting the given points into y=mx+c to form a system of linear equations and solving for m and c.

🎯 Exam Tip: For problems requiring 'm' and 'c' from two points, both the two-point formula method and the substitution method are valid. Choose the method you are most comfortable with, but always double-check your algebraic calculations, especially when solving simultaneous equations.

Question 6. The vertices of a triangle are A(3, 4), B(2, 0), and C(-1, 6). Find the equations of
(a) side BC
(b) the median AD
(c) the midpoints of sides AB and BC.
Answer:
Solution:
Vertices of \( \triangle ABC \) are A(3, 4), B(2, 0) and C(-1, 6).
(a) Equation of a line in two-point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
\( \therefore \) the equation of the side BC is
\( \frac{y-0}{6-0} = \frac{x-2}{-1-2} \) ......[B = \( (x_1, y_1) \) = (2, 0), C = \( (x_2, y_2) \) = (-1, 6)]
\( \therefore \frac{y}{6} = \frac{x-2}{-3} \)
\( \therefore \) y = -2(x - 2)
\( \therefore \) 2x + y - 4 = 0
(b) Let D be the midpoint of side BC.
Then, AD is the median through A.
\( \therefore D = (\frac{2+(-1)}{2}, \frac{0+6}{2}) = (\frac{1}{2}, 3) \)
The median AD passes through the points A(3, 4) and D\( (\frac{1}{2}, 3) \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(3, 4), B(2, 0), और C(-1, 6) हैं। इसमें भुजा BC का मध्यबिंदु D दर्शाया गया है। शीर्ष A से मध्यबिंदु D तक एक रेखा खींची गई है, जो त्रिभुज की माध्यिका (median) AD को प्रदर्शित करती है।
\( \therefore \) the equation of the median AD is
\( \frac{y-4}{3-4} = \frac{x-3}{\frac{1}{2}-3} \)
\( \therefore \frac{y-4}{-1} = \frac{x-3}{\frac{1-6}{2}} \)
\( \therefore \frac{y-4}{-1} = \frac{x-3}{-\frac{5}{2}} \)
\( \therefore \frac{1}{5}(y-4) = x-3 \)
\( \therefore \) 5y – 20 = 2x - 6
\( \therefore \) 2x – 5y + 14 = 0
(c) Let D and E be the midpoints of side AB and side BC respectively.
\( \therefore D = (\frac{3+2}{2}, \frac{4+0}{2}) = (\frac{5}{2}, 2) \) and
\( E = (\frac{2+(-1)}{2}, \frac{0+6}{2}) = (\frac{1}{2}, 3) \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है जिसके शीर्ष B(2, 0) और C(-1, 6) हैं। इसमें भुजा AB का मध्यबिंदु D और भुजा BC का मध्यबिंदु E दर्शाया गया है। बिंदुओं D और E को एक रेखाखंड से जोड़ा गया है, जो DE को प्रदर्शित करता है।
the equation of the line DE is
\( \frac{y-2}{3-2} = \frac{x-\frac{5}{2}}{\frac{1}{2}-\frac{5}{2}} \)
\( \therefore \frac{y-2}{1} = \frac{x-\frac{5}{2}}{\frac{1-5}{2}} \)
\( \therefore y-2 = \frac{x-\frac{5}{2}}{-2} \)
\( \therefore \) -4(y - 2) = 2x - 5
\( \therefore \) -4y + 8 = 2x - 5
\( \therefore \) 2x + 4y – 13 = 0
In simple words: To find the equation of a side, median, or line segment connecting midpoints of a triangle, first determine the coordinates of the relevant points (vertices or midpoints using the midpoint formula), then apply the two-point form of a linear equation using those coordinates.

🎯 Exam Tip: When dealing with triangles, remember the midpoint formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \) for finding midpoints. A median connects a vertex to the midpoint of the opposite side. Clearly label all points and show steps for midpoint calculations to avoid errors.

Question 7. Find the x and y-intercepts of the following lines:
(a) \( \frac{x}{3} + \frac{y}{2} = 1 \)
(b) \( \frac{3x}{2} + \frac{2y}{3} = 1 \)
(c) 2x – 3y + 12 = 0
Answer:
Solution:
(a) Given equation of the line is \( \frac{x}{3} + \frac{y}{2} = 1 \)
This is of the form \( \frac{x}{a} + \frac{y}{b} = 1 \),
where x-intercept = a, y-intercept = b
\( \therefore \) x-intercept = 3, y-intercept = 2
(b) Given equation of the line is \( \frac{3x}{2} + \frac{2y}{3} = 1 \)
\( \therefore \frac{x}{(\frac{2}{3})} + \frac{y}{(\frac{3}{2})} = 1 \)
This is of the form \( \frac{x}{a} + \frac{y}{b} = 1 \),
where x-intercept = a, y-intercept = b
\( \therefore \) x-intercept = \( \frac{2}{3} \) and y-intercept = \( \frac{3}{2} \)
(c) Given equation of the line is 2x – 3y + 12 = 0
\( \therefore \) 2x – 3y = -12
\( \therefore \frac{2x}{-12} - \frac{3y}{-12} = 1 \)
\( \therefore \frac{x}{-6} + \frac{y}{4} = 1 \)
This is of the form \( \frac{x}{a} + \frac{y}{b} = 1 \),
where x-intercept = a, y-intercept = b
\( \therefore \) x-intercept = -6 and y-intercept = 4
In simple words: To find the x and y-intercepts, convert the given linear equation into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), where 'a' is the x-intercept and 'b' is the y-intercept. If the equation is in standard form (Ax+By+C=0), rearrange it to equal 1 and divide by the constant term.

🎯 Exam Tip: Ensure the right-hand side of the equation is always 1 when converting to intercept form. Remember that the x-intercept is the point where the line crosses the X-axis (y=0), and the y-intercept is where it crosses the Y-axis (x=0).

Question 8. Find the equations of a line containing the point A(3, 4) and make equal intercepts on the co-ordinate axes.
Answer:
Solution:
Let the equation of the line be
\( \frac{x}{a} + \frac{y}{b} = 1 \) .....(i)
Since, the required line make equal intercepts on the co-ordinate axes.
\( \therefore \) a = b
\( \therefore \) (i) reduces to x + y = a .....(ii)
Since the line passes through A(3, 4).
3 + 4 = a
i.e. a = 7
Substituting a = 7 in (ii) to get
x + y = 7
In simple words: If a line has equal intercepts on the coordinate axes, its equation is x/a + y/a = 1, which simplifies to x + y = a. If this line also passes through a given point, substitute the point's coordinates into the simplified equation to find the value of 'a'.

🎯 Exam Tip: For lines with equal intercepts, the equation simplifies significantly. Remember that 'equal intercepts' means both x-intercept (a) and y-intercept (b) are numerically the same. Pay attention to the sign if the intercepts are in different quadrants.

Question 9. Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, -1) and C(-4, -3).
Answer:
Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(2, 5), B(6, -1), और C(-4, -3) हैं। इसमें तीनों शीर्षों से सम्मुख भुजाओं पर लंब खींचे गए हैं: AD (A से BC पर), BE (B से AC पर), और CF (C से AB पर)। ये लंब त्रिभुज की ऊँचाई (altitudes) को दर्शाते हैं।
A(2, 5), B(6, -1), C(-4, -3) are the vertices of \( \triangle ABC \).
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of \( \triangle ABC \).
Slope of BC = \( \frac{-3-(-1)}{-4-6} = \frac{-2}{-10} = \frac{1}{5} \)
\( \therefore \) slope of AD = -5 ........[ \( \therefore \) AD \( \perp \) BC]
Since, altitude AD passes through (2, 5) and has slope -5.
\( \therefore \) the equation of the altitude AD is
y - 5 = -5(x - 2)
\( \therefore \) y - 5 = -5x + 10
\( \therefore \) 5x + y - 15 = 0
Now, slope of AC = \( \frac{-3-5}{-4-2} = \frac{-8}{-6} = \frac{4}{3} \)
\( \therefore \) slope of BE = \( -\frac{3}{4} \) .......[ \( \therefore \) BE \( \perp \) AC]
Since, altitude BE passes through (6, -1) and has slope \( -\frac{3}{4} \).
\( \therefore \) the equation of the altitude BE is
y-(-1) = \( -\frac{3}{4} \)(x-6)
\( \therefore \) 4(y + 1) = -3(x - 6)
\( \therefore \) 3x + 4y – 14 = 0
Also, slope of AB = \( \frac{-1-5}{6-2} = \frac{-6}{4} = -\frac{3}{2} \)
\( \therefore \) slope of CF = \( \frac{2}{3} \) ........[ \( \therefore \) CF \( \perp \) AB]
Since, altitude CF passes through (-4, -3) and has slope \( \frac{2}{3} \).
\( \therefore \) the equation of the altitude CF is
y-(-3) = \( \frac{2}{3} \)[x-(-4)]
\( \therefore \) 3(y + 3) = 2(x + 4)
\( \therefore \) 2x – 3y – 1 = 0
In simple words: To find the equation of an altitude, first calculate the slope of the side perpendicular to it. The slope of the altitude will be the negative reciprocal of that side's slope. Then, use the point-slope form \( y-y_1 = m(x-x_1) \) with the altitude's slope and the coordinates of the corresponding vertex.

🎯 Exam Tip: The key concept for altitudes is perpendicularity. Remember that the product of slopes of two perpendicular lines is -1 (m1 * m2 = -1). Accurately calculating the slopes and applying the point-slope form for each altitude is crucial for scoring well.